You say:
Therefore, the potential energy of an object has to be absolute, a constant in some sense.
The (gravitational) potential energy is a function of position so it varies as objects move around in space. The PE at some fixed point is sort of constant in the sense that all observers will agree on it's value, but note that PE has a global gauge symmetry - you can add a constant value to it without changing the physics because only changes of PE appear in the equations of motion.
We normally take the gravitational PE to be zero in flat spacetime far from any other masses. Therefore the PE near Earth's surface is negative, and the PE near whatever passes for Jupiter's surface is considerably more negative. The (positive) kinetic energy is therefore a lot greater near Jupiter's surface to keep the total energy constant.
In the elevator scenario, the elevator frame is getting accelerated; hence, the when you draw the free-body diagram, with respect to the elevator, the pseudo force acts downwards (opposite to the direction in which the frame is getting accelerated). Hence, the apparent weight increases as the pseudo force gets added up with the weight of the person.
Suppose the acceleration of the elevator is $a$ and the mass of the body is $m$, then the apparent weight of the body in the elevator frame is -
$$
N = m(a + g)
$$
In the second scenario, the buoyant force acts in the upward direction, because the buoyant force is always directed against the pressure gradient i.e, the direction in which pressure decreases. (Much like an electric field directed in the direction where the potential decreases)
Of course, the buoyant force exerted is equal to the weight of the fluid displaced by the body (Which is the Archimedes principle); but -
Drawing the FBD in the second case yields the weight of the body acting downwards, and the buoyant force acting upwards. This results in the weight decreasing (since the buoyant force is subtracted from the weight, not added up with it), and not increasing.
Say, the buoyant force acting on the body is $B$ and the actual weight is $W$, the net weight of the body (acting in the downward direction) then would be -
$$
W' = W - B
$$
Which is why the apparent weight of the body in the liquid decreases.
(This is considering that the density of the body is greater than the density of the liquid, in the case where it is opposite (the body doesn't sink; but floats partially), the signs of $W$ and $B$ are swapped and the net force is acting in the upward direction. In another scenario where the the weight of the the body is equal to the buoyant force, the net force on the body then is zero, hence it floats being completely submerged)
Keep in mind that a body loses weight in a liquid which is equal to the weight of the liquid displaced by it/equal to the buoyant force.
As for the bonus question, look into the answer to this question -
https://physics.stackexchange.com/a/296537/134658
Best Answer
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.